Difference between revisions of "2000 AMC 12 Problems/Problem 12"
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<cmath>A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C = 125 - 13 = 112 \Rightarrow \mathrm{(E)}</cmath> | <cmath>A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C = 125 - 13 = 112 \Rightarrow \mathrm{(E)}</cmath> | ||
− | Alternatively, a quick AMC heuristic method would be to consider the equality case since that often gives a maximum or minimum. So <math>A=M=C=4</math> gives <math>AMC+AM+AC+MC = 112</math>, and that is the greatest answer choice, so the answer is <math>\ | + | Alternatively, a quick AMC heuristic method would be to consider the equality case since that often gives a maximum or minimum. So <math>A=M=C=4</math> gives <math>AMC+AM+AC+MC = 112</math>, and that is the greatest answer choice, so the answer is <math>\boxed{E}</math>. |
== See also == | == See also == |
Revision as of 06:34, 28 October 2013
Problem
Let and be nonnegative integers such that . What is the maximum value of ?
Solution
By AM-GM, ; thus is maximized at , which occurs when .
Alternatively, a quick AMC heuristic method would be to consider the equality case since that often gives a maximum or minimum. So gives , and that is the greatest answer choice, so the answer is .
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.